Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $p \neq 0$. $z = \dfrac{2p^2 + 30p + 112}{-4p^2 - 68p - 288} \div \dfrac{p + 7}{p - 1} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $z = \dfrac{2p^2 + 30p + 112}{-4p^2 - 68p - 288} \times \dfrac{p - 1}{p + 7} $ First factor out any common factors. $z = \dfrac{2(p^2 + 15p + 56)}{-4(p^2 + 17p + 72)} \times \dfrac{p - 1}{p + 7} $ Then factor the quadratic expressions. $z = \dfrac {2(p + 8)(p + 7)} {-4(p + 8)(p + 9)} \times \dfrac {p - 1} {p + 7} $ Then multiply the two numerators and multiply the two denominators. $z = \dfrac { 2(p + 8)(p + 7) \times (p - 1)} { -4(p + 8)(p + 9) \times (p + 7)} $ $z = \dfrac {2(p + 8)(p + 7)(p - 1)} {-4(p + 8)(p + 9)(p + 7)} $ Notice that $(p + 8)$ and $(p + 7)$ appear in both the numerator and denominator so we can cancel them. $z = \dfrac {2\cancel{(p + 8)}(p + 7)(p - 1)} {-4\cancel{(p + 8)}(p + 9)(p + 7)} $ We are dividing by $p + 8$ , so $p + 8 \neq 0$ Therefore, $p \neq -8$ $z = \dfrac {2\cancel{(p + 8)}\cancel{(p + 7)}(p - 1)} {-4\cancel{(p + 8)}(p + 9)\cancel{(p + 7)}} $ We are dividing by $p + 7$ , so $p + 7 \neq 0$ Therefore, $p \neq -7$ $z = \dfrac {2(p - 1)} {-4(p + 9)} $ $ z = \dfrac{-(p - 1)}{2(p + 9)}; p \neq -8; p \neq -7 $